# trans - Transformation of Stresses and Strains David...

This preview shows pages 1–3. Sign up to view the full content.

Transformation of Stresses and Strains David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 May 14, 2001 Introduction One of the most common problems in mechanics of materials involves transformation of axes. For instance, we may know the stresses acting on xy planes, but are really more interested in the stresses acting on planes oriented at, say, 30 to the x axis as seen in Fig. 1, perhaps because these are close-packed atomic planes on which sliding is prone to occur, or is the angle at which two pieces of lumber are glued together in a “scarf” joint. We seek a means to transform the stresses to these new x 0 y 0 planes. Figure 1: Rotation of axes in two dimensions. These transformations are vital in analyses of stress and strain, both because they are needed to compute critical values of these entities and also because the tensorial nature of stress and strain is most clearly seen in their transformation properties. Other entities, such as moment of inertia and curvature, also transform in a manner similar to stress and strain. All of these are second-rank tensors, an important concept that will be outlined later in this module. Direct approach The rules for stress transformations can be developed directly from considerations of static equilibrium. For illustration, consider the case of uniaxial tension shown in Fig. 2 in which all stresses other than σ y are zero. A free body diagram is then constructed in which the specimen is “cut” along the inclined plane on which the stresses, labeled σ y 0 and τ x 0 y 0 , are desired. The key here is to note that the area on which these transformed stresses act is diﬀerent than the area normal to the y axis, so that both the areas and the forces acting on them need to be “transformed.” Balancing forces in the y 0 direction (the direction normal to the inclined plane): 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Figure 2: An inclined plane in a tensile specimen. ( σ y A )cos θ = σ y 0 ± A cos θ ² σ y 0 = σ y cos 2 θ (1) Similarly, a force balance in the tangential direction gives τ x 0 y 0 = σ y sin θ cos θ (2) Example 1 Consider a unidirectionally reinforced composite ply with strengths ˆ σ 1 in the ﬁber direction, ˆ σ 2 in the transverse direction, and ˆ τ 12 in shear. As the angle θ between the ﬁber direction and an applied tensile stress σ y is increased, the stress in the ﬁber direction will decrease according to Eqn. 1. If the ply were to fail by ﬁber fracture alone, the stress σ y,b needed to cause failure would increase with misalignment according to σ y,b σ 1 / cos 2 θ . However, the shear stresses as given by Eqn. 2 increase with θ ,soth e σ y stress needed for shear failure drops. The strength σ y,b is the smaller of the stresses needed to cause ﬁber-direction or shear failure, so the strength becomes limited by shear after only a few degrees of misalignment. In fact, a 15 oﬀ-axis tensile specimen has been proposed as a means of measuring intralaminar shear strength. When
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 11/04/2011 for the course MME 512 at Miami University.

### Page1 / 15

trans - Transformation of Stresses and Strains David...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online